Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
An electron is smaller than a proton. An electron has the opposite charge of a proton. An electron is outside the nucleus, while a proton is part of the nucleus.
The change in energy of the system : -63 J
<h3>Further explanation</h3>
Given
279 J work
216 J heat
Required
The change in energy
Solution
Laws of thermodynamics 1
ΔU=Q+W
Rules :
- receives heat, Q +
- releases heat, Q -
- work is done by a system, W -
- work is done on a system, W +
a gas work on the surrounding : W =-279 J
a gas absorb heat from surrounding : Q = +216 J
Internal energy :
= -279+216
= -63 J
Answer:
He assumed wrongly
Explanation:
It happened this way because the person assumed wrongly.
Most other compounds have the same physical appearance as water especially when in liquid form.
- This can be a very tricky one in the laboratory.
- The sense of taste is not the best way to fathom what a particular compound is made up of.
- This is the reason why this student or person faced this sort of problem.
- Water has different properties that can be tested for using simple techniques in the laboratory.
The answer to this question would be:2 Mg<span> + O</span>2<span> = 2 MgO
Magnesium is a metal that can be oxidized easily. The reaction of magnesium and oxygen can be said as burning the magnesium. The reaction can provide a bright light with violet color. Magnesium used mostly in solid state and the oxygen is gas.</span>
